Answer:
adding 750 grams of water at 60° Celsius
.
Explanation:
- We can calculate the amount of heat lost from 750 grams of water at 80°C to be lowered by 10°C using the relation:
<em>Q = m.c.ΔT,</em>
Where, Q is the amount of heat lost by water (Q = ??? J).
m is the mass of water (m = 750.0 g).
c is the specific heat capacity of the water (c = 4.18 J/g.°C).
ΔT is the temperature difference (ΔT = final T - initial T = - 10.0°C, the temperature of water is lowered by 10.0°C).
∴ Q = m.c.ΔT = (750.0 g)(4.18 J/g.°C)(- 10.0°C) = - 31350.0 J = -31.350 kJ.
Now, we can calculate the Q that is gained by the different added amounts of water:
- <em>adding 750 grams of water at 50° Celsius
:</em>
ΔT = 70.0°C - 50.0°C = 20.0°C,
∴ Q = m.c.ΔT = (750.0 g)(4.18 J/g.°C)(20.0°C) = 62700.0 J = 62.70 kJ.
- <em>adding 325 grams of water at 60° Celsius
:</em>
ΔT = 70.0°C - 60.0°C = 10.0°C,
∴ Q = m.c.ΔT = (325.0 g)(4.18 J/g.°C)(10.0°C) = 13585.0 J = 13.585 kJ.
- <em>adding 750 grams of water at 60° Celsius
:</em>
ΔT = 70.0°C - 60.0°C = 10.0°C,
∴ Q = m.c.ΔT = (750.0 g)(4.18 J/g.°C)(10.0°C) = 31350.0 J = 31.350 kJ.
- <em>adding 1000 grams of water at 55° Celsius:</em>
ΔT = 70.0°C - 55.0°C = 15.0°C,
∴ Q = m.c.ΔT = (1000.0 g)(4.18 J/g.°C)(15.0°C) = 62700.0 J = 62.70 kJ.
<em>adding 750 grams of water at 60° Celsius</em>
